Let $E :=\mathbb R$ and $K:E \to E$ such that $\int_E K(x) \mathrm d x=1$. We define $$K_\otimes:E^2 \to E, (x_1, x_2) \mapsto K(x_1)K(x_2).$$ Then
$$ \int_{E^2} K_\otimes (x) \mathrm d x = \int_{E^2} K(x_1) K(x_2) \mathrm d (x_1,x_2) = \int_E \left [\int_E K(x_1) \mathrm d x_1 \right ] K(x_2)\mathrm d x_2=1. $$
Let $h>0$ and $K_h:E^2 \to E, x \mapsto h^{-2}K_\otimes(x/h)$.
Then I compute $M:=\int_{E^2} K_h (x) \mathrm d x$ in $2$ ways but obtain different results. Could you elaborate on where I got wrong? $$ M = \int_{E^2} \frac{1}{h^2}K_\otimes \left (\frac{x}{h} \right)\mathrm d x = \frac{1}{h} \int_{E^2} K_\otimes \left (u \right)\mathrm d u = \frac{1}{h}. $$
On the other hand, $$ \begin{aligned} M &= \int_{E^2} \frac{1}{h^2}K_\otimes \left (\frac{x}{h} \right)\mathrm d x \\ &= \int_{E^2} \frac{1}{h^2} K \left (\frac{x_1}{h} \right) K \left (\frac{x_2}{h} \right) \mathrm d x_1 \mathrm d x_2 \\ &= \int_{E} \left [\int_{E} \frac{1}{h} K \left (\frac{x_1}{h} \right) \mathrm d x_1 \right] \frac{1}{h} K \left (\frac{x_2}{h} \right) \mathrm d x_2 \\ &= \int_E \frac{1}{h} K \left (\frac{x_2}{h} \right) \mathrm d x_2 = 1. \end{aligned} $$
Your error is in your first computation of the double integral. There, $$ dx=h^2du $$ since both variables $x_1$ and $x_2$ are affected by the $h$ factor.